Problem: Given that $-7$ is a solution to $x^2 + bx -28 = 0$, what is the value of $b$?
The product of the roots of this quadratic is $-28/1=-28$, so the other solution must be $-28/-7=4$. That means that the sum of the solutions is $-7+4=-3$. The sum of the solutions is also $-b/1=-b$. Thus, $-b=-3$ and $b=\boxed{3}$.